Question

A student did the Synthesis of Salicylic Acid experiment starting with 3.89 g of methyl salicylate (C8H8O3) and using 40 mL of 6.0 M NaOH. To evaluate her success she needs to determine the limiting reagent. The reaction goes as follows:
C8H8O3 + H2O ==> C7H6O3 + CH3OH.
Assuming the solution of NaOH is mostly water, she uses the volume of NaOH = volume of H2O. Also she assumes that the density of water is 1.000 g/mL. Which is the limiting reagent methyl salicylate or water?
How many moles of salicylic acid can be produced if methyl salicylate is limiting?

How many moles of salicylic acid can be produced if water is limiting?

Which is the limiting reagent: methyl salicylate or water

Answer 1

Convert grams methyl salicylate to moles. moles = grams/molar mass.

Convert grams H2O to moles. same process.

Using the coefficients in the balanced equation, convert moles methyl salicylate to moles salicylic acid (SA). The coefficients are 1:1 which means moles methyl salicylate will be the same as the moles of SA. This procedure effectively tells you that if all 3.89 g methyl salicylate reacted with an excess of NaOH, then ?? moles SA would be formed.

Now do the same for the H2O, which tells you that if all of the H2O reacted with an excess of methyl salicylate, then ?? moles SA would be formed.

The two numbers probably will not agree which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that smaller value is the limiting reagent.